A5.00-m-long uniform ladder, weighing 200 n, rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.20 m away from the wall. the coefficient of static friction between the ladder and the floor is 0.200. how far up the ladder, measured along the ladder, can a 600-n person climb before the ladder begins to slip

0.488  m  

Explanation:

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N  can climb be h from the ground .

Distance from the base point  where ladder touches the floor  = h / tanθ

= h / 4.04

Total reaction force = total downward force

R = 200 + 600

800 N

Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder  where it  touches the floor  and balancing them

200 x 1.2 x .5 + 600 x   h / tanθ  = μ R x  1.2 / tanθ ( reaction  at the top point of ladder where it touches the wall is  R₁ and

R₁ =μ R   )

= 200 x 1.2 x .5 + 600 x   h / tanθ  = 160 x 1.2 / tanθ

120  - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

72.48= 148.51 h

h = 0.488  m  

=

the answer is b i think

ok

explanation: